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DISCLAIMER This problem is part of a COMPLETED competition and will not be reused
I was hoping someone could create and explain a solution efficient enough to run file sizes at most 15 x 15 tiles. Below is the problem:
Tiled Floor (40 points)
Sam recently hired a tile installer to tile a kitchen floor. The floor was supposed to be very colorful with no two
adjacent tiles having the same color. Unfortunately, the installer failed to ensure adjacent tiles have different
colors. The mortar is still wet, but it is difficult to lift just one tile at a time, so the installer is limited to swapping
adjacent tiles. The question is how to exchange adjacent tiles in as few moves as possible so that the floor meets
the criteria that no two adjacent tiles have the same color.
A sample input is
RGR
RPC
GRB
YPG
Representing the tiles on a three by four floor where R is a red tile, G is green, B is blue, C is cyan, P is purple, and
Y is yellow. In general, the input will be a series of lines of R, G, B, C, P, and Y. Each line will have the same length
and there will be a maximum of 15 lines with 15 characters in each line. The output is to be an integer
representing the number of swaps of adjacent tiles. For this problem, “adjacent” means touching and in the same
row or column; for example, the only two tiles are adjacent to the yellow tile in the lower left corner of the above
floor: the green tile at the start of the third row and the purple tile in the middle of the fourth row.
The output for the above floor will be
2
since the red tile at the start of row 2 can be swapped with the green tile at the start of row three, and then the
red tile in middle of row 3 can be swapped with the blue tile at the end. This gives the arrangement
RGR
GPC
RBR
YPG
Other fixes are possible such as exchanging the first two tiles on row 2 to get PRC and then exchanging the middle
tiles in rows 3 and 4. Your program does not print the resulting floor arrangement, just the minimum number of
tile swaps that must take place. Sometimes it is not possible to fix a floor:
GGYGP
CGGRG
This isn’t possible to tile because there are 6 Gs and a floor this size can fit only 5 without two being adjacent. In
cases where there is no solution, the output is to be
not possible
I have created a solution but it works only for approximately 16 tiles (4 x 4), any more takes an enormous amount of time. This is because of the recursive and naive nature of this function, for every call it calls itself at least 4 times.
Below is my attempted solution, keep in mind that there are extra methods from previous attempts and that minimumSwaps() is the main recursive method:
import java.util.*;
import java.io.*;
class Main {
private static ArrayList<String[][]> solutions = new ArrayList<String[][]>();
private static ArrayList<Integer> moves = new ArrayList<Integer>();
private static int min = Integer.MAX_VALUE;
public static void main(String[] args) throws Exception {
File file = new File("Tiles.txt");
Scanner scan = new Scanner(file);
Scanner scan1 = new Scanner(file);
int length = 0;
while (scan1.hasNextLine()) {
scan1.nextLine();
length++;
}
String[][] tiles = new String[length][];
for (int i = 0; i < length; i++) {
String line = scan.nextLine();
tiles[i] = new String[line.length()];
for (int l = 0; l < tiles[i].length; l++) {
tiles[i][l] = line.substring(l, l + 1);
}
}
System.out.println("Start");
minimumSwaps(tiles, 0, new ArrayList<String>());
//System.out.println(Arrays.toString(findCount(tiles)));
//findSolutions(new String[tiles.length][tiles[0].length], findCount(tiles), 0, 0);
//System.out.println(solutions.size());
System.out.println(min);
//display();
}
//tilesIDs: more efficient way to check if computer has seen previous situation
//how to know if there are moves that do not involve problem areas that reduces total number of moves?
public static void minimumSwaps (String[][] tiles, int moves, ArrayList<String> tilesIDs) {
if (moves < min) {
String newID = computeID(tiles);
if (linearSearch(tilesIDs, newID)) return;
tilesIDs.add(newID);
if (solved(tiles)) {
//Main.moves.add(moves);
if (moves < min) min = moves;
//solutions.add(cloneTiles(tiles));
}
else if (moves < min - 1) {
for (int i = 0; i < tiles.length; i++) {
for (int l = 0; l < tiles[i].length; l++) {
if (adjacentPresent(tiles, tiles[i][l], i, l)) {
try {
String[][] newTiles = cloneTiles(tiles);
String current = newTiles[i][l];
newTiles[i][l] = newTiles[i][l - 1];
newTiles[i][l - 1] = current;
minimumSwaps(newTiles, moves + 1, (ArrayList<String>)(tilesIDs.clone()));
}
catch (Exception e) {}
try {
String[][] newTiles = cloneTiles(tiles);
String current = newTiles[i][l];
newTiles[i][l] = newTiles[i][l + 1];
newTiles[i][l + 1] = current;
minimumSwaps(newTiles, moves + 1, (ArrayList<String>)(tilesIDs.clone()));
}
catch (Exception e) {}
try {
String[][] newTiles = cloneTiles(tiles);
String current = newTiles[i][l];
newTiles[i][l] = newTiles[i - 1][l];
newTiles[i - 1][l] = current;
minimumSwaps(newTiles, moves + 1, (ArrayList<String>)(tilesIDs.clone()));
}
catch (Exception e) {}
try {
String[][] newTiles = cloneTiles(tiles);
String current = newTiles[i][l];
newTiles[i][l] = newTiles[i + 1][l];
newTiles[i + 1][l] = current;
minimumSwaps(newTiles, moves + 1, (ArrayList<String>)(tilesIDs.clone()));
}
catch (Exception e) {}
}
}
}
}
}
}
public static boolean linearSearch(ArrayList<String> IDs, String newID) {
for (String ID : IDs) if (ID.equals(newID)) return true;
return false;
}
public static String computeID(String[][] tiles) {
String ID = "";
for (String[] letters : tiles) {
for (String letter : letters) {
ID += letter;
}
}
return ID;
}
public static String[][] cloneTiles(String[][] tiles) {
String[][] newTiles = new String[tiles.length][tiles[0].length];
for (int i = 0; i < tiles.length; i++) {
newTiles[i] = tiles[i].clone();
}
return newTiles;
}
public static boolean solved(String[][] tiles) {
for (int i = 0; i < tiles.length; i++) {
for (int l = 0; l < tiles[i].length; l++) {
if (adjacentPresent(tiles, tiles[i][l], i, l)) return false;
}
}
return true;
}
public static int minMoves() {
int min = Integer.MAX_VALUE;
for (int num : moves) if (num < min) min = num;
return min;
}
public static void findSolutions(String[][] tiles, int[] count, int i, int l) {
String[] colors = {"R", "G", "B", "C", "P", "Y"};
for (int z = 0; z < 6; z++) {
//System.out.println("testing");
if (!adjacentPresent(tiles, colors[z], i, l) && count[z] > 0) {
String[][] newTiles = new String[tiles.length][tiles[0].length];
for (int a = 0; a < newTiles.length; a++) {
for (int b = 0; b < newTiles[0].length; b++) {
newTiles[a][b] = tiles[a][b]; // clone does not work properly?
}
}
newTiles[i][l] = colors[z];
//System.out.println(Arrays.deepToString(newTiles));
int[] newCount = count.clone();
newCount[z]--;
if (l == tiles[0].length - 1 && i != tiles.length - 1) {
findSolutions(newTiles, newCount, i + 1, 0);
}
else if (l < tiles[0].length - 1) {
findSolutions(newTiles, newCount, i, l + 1);
}
else if (l == tiles[0].length - 1 && i == tiles.length - 1) {
solutions.add(newTiles);
}
}
}
}
public static boolean adjacentPresent(String[][] tiles, String color, int i, int l) {
try {
if (tiles[i][l + 1].equals(color)) return true;
}
catch (Exception e) {}
try {
if (tiles[i][l - 1].equals(color)) return true;
}
catch (Exception e) {}
try {
if (tiles[i + 1][l].equals(color)) return true;
}
catch (Exception e) {}
try {
if (tiles[i - 1][l].equals(color)) return true;
}
catch (Exception e) {}
return false;
}
public static int[] findCount(String[][] tiles) {
int[] count = new int[6];
for (String[] line : tiles) {
for (String letter : line) {
switch (letter) {
case "R": count[0]++;
break;
case "G": count[1]++;
break;
case "B": count[2]++;
break;
case "C": count[3]++;
break;
case "P": count[4]++;
break;
case "Y": count[5]++;
break;
}
}
}
return count;
}
public static void display() {
for (String[][] lines : solutions) {
for (String[] line : lines) {
for (String letter : line) {
System.out.print(letter);
}
System.out.println();
}
System.out.println("\n\n");
}
}
}
Improving the algorithm
A breadth-first search would yield, as first result, an optimal solution. It could still be slow on larger problems where the solution is deeper in; or in the worst case, when there is no solution at all.
Your current algorithm looks like backtracking, which is depth-first, and therefore you need to look at all possible solutions before being sure you have found the shortest one. This is very wasteful.
Improving the data representation
You have 6 colors in a 15x15 grid. You are currently storing up to 15x15=225 strings, and constantly copying that String[][] over. It would be a lot more efficient to use a single byte[] (of length dim x dim), which can be copied over faster. Use integers (1, 2, ...) instead of color-chars ("R", "Y", ...). With a single dimension you have to do some math to check for adjacency, but it is nothing too fancy; and you win a lot of memory locality.
I am teaching myself java using the cs106a course from Stanford.
Currently I am on chapter 10 of the book "The Art and Science of Java".
The problem is to write a 3x3 Magic Square.
The exercise:
You have to write a 3x3 array
Each side of the array(Magic Square) has to equal 15
The problem:
The program I wrote works, the assignment is complete, this question is for self learning. As a beginner I would like to improve the method SumOfSides() and make it smaller and more efficient. I tried iterating the array in this method but still have issues. Is there a way to make it more efficient?
public void run() {
//set the font
setFont("Helvetica-40");
//fill the array
fillArray();
//sum up all sides
SumOfSides();
//check if all of the sides in the magic square array equal 15:
checkSides(mSqr);
//I used this for debugging purposes only:
//showSides();
}
//for debugging purposes:
public void showSides() {
println(sumRight0);
println(sumRight1);
println(sumRight2);
println(sumBottom0);
println(sumBottom1);
println(sumBottom2);
println(sumDiagonalUp);
println(sumDiagonalDown);
}
public void SumOfSides() {
sumRight0 = mSqr[0][0] + mSqr[0][1] + mSqr[0][2];
sumRight1 = mSqr[1][0] + mSqr[1][1] + mSqr[1][2];
sumRight2 = mSqr[2][0] + mSqr[2][1] + mSqr[2][2];
sumBottom0 =mSqr[0][0] + mSqr[1][0] + mSqr[2][0];
sumBottom1 =mSqr[0][1] + mSqr[1][1] + mSqr[2][1];
sumBottom2 =mSqr[0][2] + mSqr[1][2] + mSqr[2][2];
sumDiagonalUp = mSqr[2][0] + mSqr[1][1]+ mSqr[0][2];
sumDiagonalDown = mSqr[0][0] + mSqr[1][1] + mSqr[2][2];
}
/*This predicate method checks if the sides
of the array add up to 15: */
public boolean checkSides(int[][] myArray) {
if (sumRight0 ==15 && sumRight1 ==15&& sumRight2==15 && sumBottom0==15&& sumBottom1==15&&
sumBottom2==15&& sumDiagonalUp==15&&sumDiagonalDown==15) {
println("True, this is a Magic Square");
return true;
} else {
println("False, the sides do not equal 15");
return false;
}
}
public void fillArray() {
int num =0;
for(int row=0; row <3; row++) {
for (int col=0; col<3; col++) {
num=readInt("");
mSqr[row][col]=num;
}
}
/*Test array values here to see
* if they were entered correctly.
*/
//println(mSqr[1][2]); //should be 6
//println(mSqr[2][0]); //should be 7
}
//instance variables:
int[][] mSqr= new int[3][3];
List<List<Integer>> new1 = new ArrayList<>();
private int sumRight0;
private int sumRight1;
private int sumRight2;
private int sumBottom0;
private int sumBottom1;
private int sumBottom2;
private int sumDiagonalUp;
private int sumDiagonalDown;
}
Perhaps the only thing is readability. You could take the values and move them into more readable variables:
int topLeft = mSqr[0][0];
int topMid = mSqr[0][1];
...
int sumLeft = topLeft + midLeft + bottomLeft;
int sumRight = topRight = midRight + bottomRight;
...
To address your concern of making it smaller, I would argue that converting the sums into loops, as you mentioned, is certainly not worth it in the case that you are doing 6 sums of 3 values each. Furthermore, each term of each sum is common to either one or two other sums, which does not provide much overlap. If you were performing larger sums (larger in number of terms in the sum, not the total value), then perhaps it would be worth it on a readability/SLOC argument.
Suppose you did want to do a loop still though, you could do something like
sumLeft = 0;
sumRight = 0;
sumTop = 0;
sumBottom = 0;
sumDiagonalUp = 0;
sumDiagonalDown = 0;
for(int i = 0; i < mSqr.length; i++) {
for(int j = 0; j < mSqr[i].length; j++) {
if (i == 0) {
sumLeft += mSqr[i][j];
}
if (i == mSqr.length - 1) {
sumRight += mSqr[i][j];
}
if (j == 0) {
sumTop += mSqr[i][j];
}
if (j == mSqr[i].length) {
sumBottom += mSqr[i][j];
}
if (i == j) {
sumDiagonalDown += mSqr[i][j];
}
if (i + j == mSqr.length - 1) {
sumDiagonalUp += mSqr[i][j];
}
}
}
The loops only provide benefit on large magic squares.
Also, I am confused by your description contrasted with your implementation. It seems you are summing each row and column, and the two diagonals of the square, as opposed to the 4 sides and the diagonals.
First, sorry for my bad english, especially for programming words, english isnt my first language.
So, i have programmed a software that detect all continuous sprites on an image and list their palettes.
Full explanation of the software here: https://www.vg-resource.com/thread-33373.html
It works alright, however, if the sprite has at least 4300-ish pixels, a stackoverflow exception is thrown.
In order to find the boundaries of every sprite, first i find the first pixel in the sheet that isnt the color of the background. Then, i start my recursive method that verify every adjacent pixels to see if they dont have the background color, then call itself on that position, their position being recorded in a boolean matrix.
The recursive method:
//pixelPosition > the position found of the current sprite, posX/posY > position of the current pixel being examined
public boolean[][] moveToNextPixel(boolean[][] pixelPosition, int posX, int posY)
{
pixelPosition[posX][posY] = true;
//If the next position isnt outside of the boundaries of the image AND if it hasnt already been recorded
// AND if it isnt the color of the background, move to that position.
if(posX + 1 < pixelPosition.length)
{
if(!pixelPosition[posX+1][posY] && !panBackgroundColor.isColorPresentInPalette(workingImage.getRGB(posX+1,posY)) )
{
moveToNextPixel(pixelPosition,posX+1,posY);
}
}
if(posX - 1 >= 0)
{
if(!pixelPosition[posX-1][posY] && !panBackgroundColor.isColorPresentInPalette(workingImage.getRGB(posX-1,posY)))
{
moveToNextPixel(pixelPosition,posX-1,posY);
}
}
if(posY + 1 < pixelPosition[0].length)
{
if(!pixelPosition[posX][posY+1] && !panBackgroundColor.isColorPresentInPalette(workingImage.getRGB(posX,posY+1)))
{
moveToNextPixel(pixelPosition,posX,posY+1);
}
}
if(posY - 1 >= 0)
{
if(!pixelPosition[posX][posY-1] && !panBackgroundColor.isColorPresentInPalette(workingImage.getRGB(posX,posY-1)))
{
moveToNextPixel(pixelPosition,posX,posY-1);
}
}
return pixelPosition;
}
//the method isColorPresentInPalette(int) check if the color in entry is in the background colors
public boolean isColorPresentInPalette( int colorRgb)
{
boolean result = false;
for( int i =0; i< backgroundPalette.length && !result;i++)
{
if(backgroundPalette[i] != null)
{
if(backgroundPalette[i].getRGB() == colorRgb)
{
result = true;
}
}
}
return result;
}
Also, if i load a sheet with normal-sized sprite first, and then load one with a huge sprite (4400+ pixels), it doesnt do the stackoverflow error... So, in the end, im pretty confused on what is the problem exactly.
So, is a recursive method really the right way for this kind of problem? If so what could i do to fix this? Otherwise, anyone see a way to determine each individuals continuous sprites and their positions?
EDITED: Originally I posted a recursive solution but didn't realize that you were doing that. I think after reading more carefully, it seems Recursion might not be the best since you will be adding so many calls given 4300 pixels.
I would just do DFS in memory in this case then. Alternatively, you might try BFS (which will search outwards from the center).
An example of DFS in memory. This basically does the same thing as the recursion above except instead of storing things on the callstack which has a limited buffer size, you would be storing memory:
import java.util.ArrayList;
import java.util.List;
import java.util.Objects;
import java.util.Stack;
public class FindNeedleInHaystack {
String[][] haystack;
class Coordinate {
int x;
int y;
public Coordinate(int x, int y) {
this.x = x;
this.y = y;
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Coordinate that = (Coordinate) o;
return x == that.x &&
y == that.y;
}
#Override
public int hashCode() {
return Objects.hash(x, y);
}
}
public FindNeedleInHaystack() {
this.haystack = new String[10][10];
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
this.haystack[i][j] = "";
}
}
}
public void addNeedle(int a_x, int a_y) {
this.haystack[a_y][a_x] = "needle";
}
public boolean hasNeedle() {
boolean[][] visited = new boolean[10][10];
return hasNeedleHelper(0, 0);
}
private List<Coordinate> neighbors(Coordinate coord, boolean[][] visited) {
List<Coordinate> neighbors = new ArrayList<>();
int x = coord.x;
int y = coord.y;
if (y + 1 < 10 && !visited[y+1][x]) neighbors.add(new Coordinate(x, y+1));
if (y - 1 >= 0 && !visited[y-1][x]) neighbors.add(new Coordinate(x, y-1));
if (x + 1 < 10 && !visited[y][x+1]) neighbors.add(new Coordinate(x + 1, y));
if (x - 1 >= 0 && !visited[y][x-1]) neighbors.add(new Coordinate(x - 1, y));
return neighbors;
}
private boolean hasNeedleHelper(int x, int y) {
Stack<Coordinate> fringe = new Stack<>();
boolean[][] visited = new boolean[10][10];
fringe.push(new Coordinate(x, y));
while(!fringe.isEmpty()) {
Coordinate toVisit = fringe.pop();
if (this.haystack[toVisit.y][toVisit.x].equals("needle")) {
return true;
} else {
visited[toVisit.y][toVisit.x] = true;
for(Coordinate coord : this.neighbors(toVisit, visited)) {
fringe.push(coord);
}
}
}
return false;
}
public static void main(String...args) {
FindNeedleInHaystack hasNeedle = new FindNeedleInHaystack();
hasNeedle.addNeedle(3, 4);
System.out.println("Has a needle?: " + hasNeedle.hasNeedle());
FindNeedleInHaystack doesntHaveNeedle = new FindNeedleInHaystack();
System.out.println("Has a needle?: " + doesntHaveNeedle.hasNeedle());
}
}
Hi I have the following method. What it does is it finds all the possible paths from the top left to bottom right of a N x M matrix. I was wondering what is the best way to optimize it for speed as it is a little slow right now. The resulted paths are then stored in a set.
EDIT I forgot to clarify you can only move down or right to an adjacent spot, no diagonals from your current position
For example
ABC
DEF
GHI
A path from the top left to bottom right would be ADEFI
static public void printPaths (String tempString, int i, int j, int m, int n, char [][] arr, HashSet<String> palindrome) {
String newString = tempString + arr[i][j];
if (i == m -1 && j == n-1) {
palindrome.add(newString);
return;
}
//right
if (j+1 < n) {
printPaths (newString, i, j+1, m, n, arr, palindrome);
}
//down
if (i+1 < m) {
printPaths (newString, i+1, j, m, n, arr, palindrome);
}
}
EDIT Here is the entirety of the code
public class palpath {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new FileReader("palpath.in"));
PrintWriter pw = new PrintWriter(new BufferedWriter(new FileWriter("palpath.out")));
StringTokenizer st = new StringTokenizer(br.readLine());
int d = Integer.parseInt(st.nextToken());
char[][] grid = new char [d][d];
String index = null;
for(int i = 0; i < d; i++)
{
String temp = br.readLine();
index = index + temp;
for(int j = 0; j < d; j++)
{
grid[i][j] = temp.charAt(j);
}
}
br.close();
int counter = 0;
HashSet<String> set = new HashSet<String>();
printPaths ("", 0, 0, grid.length, grid[0].length, grid, set);
Iterator<String> it = set.iterator();
while(it.hasNext()){
String temp = it.next();
StringBuilder sb = new StringBuilder(temp).reverse();
if(temp.equals(sb.toString())) {
counter++;
}
}
pw.println(counter);
pw.close();
}
static public void printPaths (String tempString, int i, int j, int m, int n, char [][] arr, HashSet<String> palindrome) {
String newString = tempString + arr[i][j];
if (i == m -1 && j == n-1) {
palindrome.add(newString);
return;
}
//right
if (j+1 < n) {
printPaths (newString, i, j+1, m, n, arr, palindrome);
}
//down
if (i+1 < m) {
printPaths (newString, i+1, j, m, n, arr, palindrome);
}
}
Given a graph of length M x N, all paths from (0,0) to (M-1, N-1) that only involve rightward and downward moves are guaranteed to contain exactly M-1 moves rightward and N-1 moves downward.
This presents us with an interesting property: we can represent a path from (0,0) to (M-1, N-1) as a binary string (0 indicating a rightward move and 1 indicating a downward move).
So, the question becomes: how fast can we print out a list of permutations of that bit string?
Pretty fast.
public static void printPaths(char[][] arr) {
/* Get Smallest Bitstring (e.g. 0000...111) */
long current = 0;
for (int i = 0; i < arr.length - 1; i++) {
current <<= 1;
current |= 1;
}
/* Get Largest Bitstring (e.g. 111...0000) */
long last = current;
for (int i = 0; i < arr[0].length - 1; i++) {
last <<= 1;
}
while (current <= last) {
/* Print Path */
int x = 0, y = 0;
long tmp = current;
StringBuilder sb = new StringBuilder(arr.length + arr[0].length);
while (x < arr.length && y < arr[0].length) {
sb.append(arr[x][y]);
if ((tmp & 1) == 1) {
x++;
} else {
y++;
}
tmp >>= 1;
}
System.out.println(sb.toString());
/* Get Next Permutation */
tmp = (current | (current - 1)) + 1;
current = tmp | ((((tmp & -tmp) / (current & -current)) >> 1) - 1);
}
}
You spend a lot of time on string memory management.
Are strings in Java mutable? If you can change chars inside string, then set length of string as n+m, and use this the only string, setting (i+j)th char at every iteration. If they are not mutable, use array of char or something similar, and transform it to string at the end
For a given size N×M of the array all your paths have N+M+1 items (N+M steps), so the first step of optimization is getting rid of recursion, allocating an array and running the recursion with while on explicit stack.
Each partial path can be extended with one or two steps: right or down. So you can easily make an explicit stack with positions visited and a step taken on each position. Put the position (0,0) to the stack with phase (step taken) 'none', then:
while stack not empty {
if stack is full /* reached lower-right corner, path complete */ {
print the path;
pop;
}
else if stack.top.phase == none {
stack.top.phase = right;
try push right-neighbor with phase none;
}
else if stack.top.phase == right {
stack.top.phase = down;
try push down-neighbor with phase none;
}
else /* stack.top.phase == down */ {
pop;
}
}
If you make a few observations about your requirements you can optimise this drastically.
There will be exactly (r-1)+(c-1) steps (where r = rows and c = columns).
There will be exactly (c-1) steps to the right and (r-1) steps down.
You therefore can use numbers where a zero bit could (arbitrarily) indicate a down step while a 1 bit steps across. We can then merely iterate over all numbers of (r-1)+(c-1) bits containing just (c-1) bits set. There's a good algorithm for that at the Stanford BitTwiddling site Compute the lexicographically next bit permutation.
First a BitPatternIterator I have used before. You could pull out the code in hasNext if you wish.
/**
* Iterates all bit patterns containing the specified number of bits.
*
* See "Compute the lexicographically next bit permutation" http://graphics.stanford.edu/~seander/bithacks.html#NextBitPermutation
*
* #author OldCurmudgeon
*/
public static class BitPattern implements Iterable<BigInteger> {
// Useful stuff.
private static final BigInteger ONE = BigInteger.ONE;
private static final BigInteger TWO = ONE.add(ONE);
// How many bits to work with.
private final int bits;
// Value to stop at. 2^max_bits.
private final BigInteger stop;
// All patterns of that many bits up to the specified number of bits.
public BitPattern(int bits, int max) {
this.bits = bits;
this.stop = TWO.pow(max);
}
#Override
public Iterator<BigInteger> iterator() {
return new BitPatternIterator();
}
/*
* From the link:
*
* Suppose we have a pattern of N bits set to 1 in an integer and
* we want the next permutation of N 1 bits in a lexicographical sense.
*
* For example, if N is 3 and the bit pattern is 00010011, the next patterns would be
* 00010101, 00010110, 00011001,
* 00011010, 00011100, 00100011,
* and so forth.
*
* The following is a fast way to compute the next permutation.
*/
private class BitPatternIterator implements Iterator<BigInteger> {
// Next to deliver - initially 2^n - 1 - i.e. first n bits set to 1.
BigInteger next = TWO.pow(bits).subtract(ONE);
// The last one we delivered.
BigInteger last;
#Override
public boolean hasNext() {
if (next == null) {
// Next one!
// t gets v's least significant 0 bits set to 1
// unsigned int t = v | (v - 1);
BigInteger t = last.or(last.subtract(BigInteger.ONE));
// Silly optimisation.
BigInteger notT = t.not();
// Next set to 1 the most significant bit to change,
// set to 0 the least significant ones, and add the necessary 1 bits.
// w = (t + 1) | (((~t & -~t) - 1) >> (__builtin_ctz(v) + 1));
// The __builtin_ctz(v) GNU C compiler intrinsic for x86 CPUs returns the number of trailing zeros.
next = t.add(ONE).or(notT.and(notT.negate()).subtract(ONE).shiftRight(last.getLowestSetBit() + 1));
if (next.compareTo(stop) >= 0) {
// Dont go there.
next = null;
}
}
return next != null;
}
#Override
public BigInteger next() {
last = hasNext() ? next : null;
next = null;
return last;
}
#Override
public void remove() {
throw new UnsupportedOperationException("Not supported.");
}
#Override
public String toString() {
return next != null ? next.toString(2) : last != null ? last.toString(2) : "";
}
}
}
Using that to iterate your solution:
public void allRoutes(char[][] grid) {
int rows = grid.length;
int cols = grid[0].length;
BitPattern p = new BitPattern(rows - 1, cols + rows - 2);
for (BigInteger b : p) {
//System.out.println(b.toString(2));
/**
* Walk all bits, taking a step right/down depending on it's set/clear.
*/
int x = 0;
int y = 0;
StringBuilder s = new StringBuilder(rows + cols);
for (int i = 0; i < rows + cols - 2; i++) {
s.append(grid[y][x]);
if (b.testBit(i)) {
y += 1;
} else {
x += 1;
}
}
s.append(grid[y][x]);
// That's a solution.
System.out.println("\t" + s);
}
}
public void test() {
char[][] grid = {{'A', 'B', 'C'}, {'D', 'E', 'F'}, {'G', 'H', 'I'}};
allRoutes(grid);
char[][] grid2 = {{'A', 'B', 'C'}, {'D', 'E', 'F'}, {'G', 'H', 'I'}, {'J', 'K', 'L'}};
allRoutes(grid2);
}
printing
ADGHI
ADEHI
ABEHI
ADEFI
ABEFI
ABCFI
ADGJKL
ADGHKL
ADEHKL
ABEHKL
ADGHIL
ADEHIL
ABEHIL
ADEFIL
ABEFIL
ABCFIL
which - to my mind - looks right.
There's an error in the logic of what I've build at the moment.
What should be happening is that my code should display a grid of 0's and 1's.
Like so:
001001
101101
010110
110010
001101
So what has to happen here is that:
For each row there can't be more than 2 numbers of the same type consecutively
the numbers are picked randomly
for each column there can't be more than 2 numbers of the same type consecutively
there can be a maximum of 3 of each type of number going by column or row
edit: to further clarify
ok so I have a row like this:
0 1 0 1 1 0
- As you can see there will always be 3 x 1, and 3 x 0
- the order of numbers is picked randomly (so it might go 0 1, or 1 1, or 0 0 to start etc)
- there can never be more than 2 numbers of the same type consecutively, for instance if it's 001100, you can see that there were 2 0's, then it had to display a 1, but then there were 2 1's, so it had to display an 0. So 011100 couldn't happen (3 1's consecutively) or 000101 (3 0's consecutively)
Based upon this, but for now not essential, the same no 2 numbers consecutively must apply in columns (so in my successful example it goes 001001 across, there are at most 2 0's consecutively. But looking down you get 010101 (that is to say, once again, no more than 2 consecutively)
So my code is as follows:
import java.util.Random;
public class Main {
public static void main(String[] args) {
int l = 6;
int w = 6;
Random rd = new Random();
// Create a grid that is 6 x 6
int[][] grid = new int[l][w];
// for each row
for (int i = 0; i < l; i++) {
int zCount = 0;
int oCount = 0;
int current;
int lastA = 2;
int lastB = 2;
// for each item in the row
for (int j = 0; j < w; j++) {
// set the current item to either 0 or 1
current = rd.nextInt(2);
// make sure there aren't already (e.g. 3 items out of 6)
// items in the row
if (j % 2 == 1) {
// hold every second element
lastA = current;
} else {
// hold every first element
lastB = current;
}
if (current == 1) {
if (oCount != 3) {
if (lastA != lastB) {
// if the two previous items aren't the same
grid[i][j] = current;
// add to the counter
oCount++;
}
}
}
if (current == 0) {
if (zCount != 3) {
if (lastA != lastB) {
// if the two previous items aren't the same
grid[i][j] = current;
// add to the counter
zCount++;
}
}
}
System.out.print(grid[i][j]);
}
System.out.println(" ");
}
}
}
The problem is it generates as follows:
010010
100001
100010
000010
100001
001000
So obviously it doesn't conform to the first, third or fourth points.
I have absolutely no idea why! Except for the columns (third point) which I haven't initialised.
Can anybody work out what the logical failure is in my code?
Thanks for your help!
Here is my procedural solution which tries to keep the amount of required code as small as possible. It is capable of computing 2D-Arrays with arbitrary rows and columns like [6, 6] or [4, 7] or [3, 8] for example. The complexity of the algorithm is O(n) with n = rows * columns.
The program computes an arbitrary 2D-Array (grid) populated with either a 0 or 1. The grid guarantees the following characteristics, formulated mathematically:
∀ r,c ∈ Integer | 0 ≤ r < grid.rows, 0 ≤ c < grid.columns :
r - 2 ≥ 0 ⇒ cardinality( distinct( grid[r][c], grid[r-1][c], grid[r-2][c] )) = 2
r + 2 < grid.rows ⇒ cardinality( distinct( grid[r][c], grid[r+1][c], grid[r+2][c] )) = 2
c - 2 ≥ 0 ⇒ cardinality( distinct( grid[r][c], grid[r][c-1], grid[r][c-2] )) = 2
c + 2 < grid.columns ⇒ cardinality( distinct( grid[r][c], grid[r][c+1], grid[r][c+2] )) = 2
or in other words:
the grid does neither contain a row nor a column which has three or more consecutive 0's or 1's.
Below the Java code I will explain how the algorithm works and why it is designed as it is:
public static void main(String[] args) {
int[][] grid = anyGrid(8, 13);
}
private static int[][] anyGrid(int rows, int cols) {
int[][] grid = new int[rows][cols];
int row = 0;
for (int col = 0; col - row < cols; col++) {
for (int r = row; r >= 0 && col - r < cols;) {
setBit(grid, r, col - r--);
}
if (row < rows - 1) row++;
}
return grid;
}
private static void setBit(int[][] grid, int row, int col) {
int vInd = calcVerticalIndicator(grid, row, col);
int hInd = calcHorizontalIndicator(grid, row, col);
if (isPartiallyRestricted(vInd, hInd)) {
grid[row][col] = flip(vInd);
} else if (isFullyRestricted(vInd, hInd)) {
grid[row][col] = vInd;
grid[row - 1][col] = flip(vInd);
} else {
grid[row][col] = Math.abs(vInd) <= 1
? flip(vInd)
: Math.abs(hInd) <= 1 ? flip(hInd) : anyBit();
}
}
private static boolean isPartiallyRestricted(int vInd, int hInd) {
return vInd == hInd;
}
private static boolean isFullyRestricted(int vInd, int hInd) {
return vInd + hInd == 1;
}
private static int calcVerticalIndicator(int[][] grid, int row, int col) {
return calcIndicator(grid, row - 1, col, row - 2, col, 2);
}
private static int calcHorizontalIndicator(int[][] grid, int row, int col) {
return calcIndicator(grid, row, col - 1, row, col - 2, 4);
}
private static int calcIndicator(int[][] grid, int row1, int col1, int row2, int col2, int unrestricted) {
try {
return grid[row1][col1] * grid[row2][col2] + (grid[row1][col1] - grid[row2][col2]) * unrestricted;
} catch (IndexOutOfBoundsException e) {
return unrestricted;
}
}
private static int anyBit() {
return (int) (Math.random() * 2);
}
private static int flip(int bit) {
return bit == 0 ? 1 : 0;
}
The challenge we face is not to ensure that there are no three consecutive 0's or 1's in a row only or in a column only. The challenge is to ensure that no three consecutive 0's or 1's are neither in a row nor in a column by providing an efficient algorithm.
The tricky situation we may run into looks like this:
Let's consider the situation where all the cells at the top and to the left of the cell outlined in blue are already populated and do not violate the rules define above.
picture a) we want to populate the cell having a blue outline. The two cells at it's top are populated with two 0's while the cells at it's left are populated with two 1's. Which value should we choose? Due to symmetry it doesn't matter if we choose a 0 or a 1. Hence, let's go with a 0.
picture b) populating the cell outlined in blue with a 0 violates one rule defined above: the grid does not contain a column with three or more consecutive 0's or 1's. Hence we have to change the value of one of the two cells above of the blue cell.
picture c) say we change the value of the cell which is immediately above the blue cell, from 0 to 1. This could result in the violation of some rules, caused by the already populated cells to the left of the modified cell.
picture d) but a violation would mean that both cells to the left must have a value of 1.
picture e) this would imply that both cells to their top must have a value of 0 which is a contradiction to a situation we assumed. Therefore, changing the cell immediately at the top of the cell outlined in blue will not cause any violation of the rules.
To address the precondition, that no cells to the right of the modified cell are already populated, the algorithm populates the grid in a diagonal way. The population of cells occur in the order as shown below:
The final thing I like to explain is how the algorithm decides which values are available to choose from for each cell. For each cell it inspects the two top-most and two left-most cells and calculates an indication value. This value is used to determine the possible values for a cell by using arithmetic calculation as follows:
if the two cells inspected are both populated with 0's return an indicator value of 0.
if the two cells inspected are both populated with 1's return an indicator value of 1.
I have selected those two values because they communicate the fact, that this values are not permitted, in an intuitive way.
Then I selected a function to communicate if both, the column cells and the row cells, restrict the cell to populate by the same value. This is the case if both indicator values are equal. Keep this characteristic in mind, because we have to find values for the situation when no restriction applies from the column cells or the row cells.
If both indicators restrict the value to populate the cell with by a different value, the sum of them is 1. This is the second characteristic we have to keep in mind when searching for proper indicator values when no restriction applies.
The last thing the algorithm has to achieve is to find proper values when no restriction applies without compromising the unique indicators defined above.
Preserving the indication when the cell is restricted by the same value can be achieved by selecting values for the row and column indicators which are different from 0 and 1 and different from each other.
Preserving the indication when the cell is restricted by both values can be achieved by selecting values being greater than 1 and having a delta to each other of at least 2.
The algorithm does indicate no restriction for a row by the values 2 and -2 and for a column by the values 4 and -4. This values do not conflict with the operations used to identify the other two cases.
I hope this documentation helps to understand the whole program and how it does solve the problem statement. I am glad to hear your comments.
Many of the solutions given are extremely long and complicated. Here's a solution with very minimal code (Ideone Example here):
int row, col, n = 8;
int[][] grid = new int[n][n], cCount = new int[n][2], rCount = new int[n][2];
Deque<Entry<Integer,Integer>> freeInd = new ArrayDeque<Entry<Integer,Integer>>();
Random rand=new Random();
for(int i = 0; i < grid.length; i++){
for(int j = 0; j < grid[0].length; j++){
// Calcualte constraints: row, col = {-1, 0, 1}, -1 => no constraint.
row = j > 1 && grid[i][j-2] == grid[i][j-1] ? (grid[i][j-1] == 0 ? 1:0):
(rCount[i][0] >= n/2 ? 1: // too many 0's
(rCount[i][1] >= n/2 ? 0:-1)); // too many 1's
col = i > 1 && grid[i-2][j] == grid[i-1][j] ? (grid[i-1][j] == 0 ? 1:0):
(cCount[j][0] >= n/2 ? 1: // too many 0's
(cCount[j][1] >= n/2 ? 0:-1)); // too many 1's
grid[i][j] = row == -1 && col == -1 ? rand.nextInt(2):(row > -1 ? row:col);
// Handle Constraints
if( row == -1 && col == -1){ // no constraint
freeInd.push(new SimpleEntry<Integer,Integer>(i, j)); // add to free indices
} else if( (row > -1 && col > -1 && row != col) // constraint conflict
|| (row > -1 && rCount[i][row] >= n/2) // count conflict
|| (col > -1 && cCount[j][col] >= n/2)){ // count conflict
Entry<Integer, Integer> last = freeInd.pop(); // grab last free index
while(i > last.getKey() || j > last.getValue()){
j = (j-1+ n)%n; // step indices back
i = (j == n-1) ? i-1:i;
rCount[i][grid[i][j]]--; // reduce counters
cCount[j][grid[i][j]]--;
}
grid[i][j] = grid[i][j] == 0 ? 1:0; // flip value
}
rCount[i][grid[i][j]]++; // increment counters
cCount[j][grid[i][j]]++;
}
}
The idea here is that you walk along each row of the matrix adding 0's and 1's abiding by the following rules:
If the current index is unconstrained (i.e. it can be 0 or 1) we choose a value randomly.
If the current index is constrained we force it to have the constrained value.
If there are multiple constraints that do not agree, we revert back to the last unconstrained index (freeInd) by first incrementally stepping backwards along the rows of the matrix, decrementing the count for the given value (0 or 1). E.g. this is done for rows with rCount[i][grid[i][j]]--. When the unconstrained vertex is finally reached, flip it's value.
Finally, increment the count of the value (0 or 1) for the current row and column. E.g. this is done for rows with rCount[i][grid[i][j]]++
The 1st problem which i found in your solution is it's initializing the value of counter value (ocount and zcount) as zero and the only way grid(array) is assigned a value is when if it's greater than three, and the way i see if i am not mistaken the value of counter is incremented in the loop in which they are checked to be greater than 3, and that condition can never be reached .
To solve this problem use the algo of backtracking by assigning the new value to a different value if the calue
A working code in jsFiddle (for 6x6 grids):
$(function(){
function print(str){
$("body").append(str + "<br/>");
}
function toBin(num, length){
if(!length){
length = 3;
}
var str = num.toString(2);
while(str.length < length){
str = 0 + str;
}
return str;
}
var wrongAnds = [
parseInt('000000111', 2),
parseInt('000111000', 2),
parseInt('111000000', 2),
parseInt('100100100', 2),
parseInt('010010010', 2),
parseInt('001001001', 2),
];
var wrongOrs = [
parseInt('111111000', 2),
parseInt('111000111', 2),
parseInt('000111111', 2),
parseInt('011011011', 2),
parseInt('101101101', 2),
parseInt('110110110', 2),
];
function test(mask){
for (var i = 0; i < 6; i++) {
if((wrongAnds[i] & mask) == wrongAnds[i]){
return false;
}
if((wrongOrs[i] | mask) == wrongOrs[i]){
return false;
}
}
return true;
}
var threeGrid = [];
var toRight = [];
var toBottom = [];
for(var mask = 1<<9-1; mask >= 0; mask--){
if(test(mask)){
threeGrid.push(mask);
}
}
function numberOfSetBits(i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}
function getCol(grid, col){
var ret = 0;
for(var i=0; i<3; i++){
ret += (grid & (1 << (i*3+col))) >> (i*2+col);
}
return ret;
}
var wrongAnds6 = [
parseInt('011100', 2),
parseInt('001110', 2)
];
var wrongOrs6 = [
parseInt('100011', 2),
parseInt('110001', 2)
];
for(var i = 0; i < threeGrid.length; i++){
for(var j = 0; j < threeGrid.length; j++){
var grid1 = threeGrid[i];
var grid2 = threeGrid[j];
var toRightOk = true;
var toBottomOk = true;
var printit = (i==0);
for(var k=0;k<3;k++){
var row = ((grid1 & wrongAnds[k]) << 3 >> (k*3)) + ((grid2 & wrongAnds[k]) >> (k*3));
var col = ((getCol(grid1, k)) << 3) + ((getCol(grid2, k)));
if(numberOfSetBits(row) != 3
|| ((wrongAnds6[0] & row) == wrongAnds6[0])
|| ((wrongAnds6[1] & row) == wrongAnds6[1])
|| ((wrongOrs6[0] | row) == wrongOrs6[0])
|| ((wrongOrs6[1] | row) == wrongOrs6[1])
) {
toRightOk = false;
}
if(numberOfSetBits(col) != 3
|| ((wrongAnds6[0] & col) == wrongAnds6[0])
|| ((wrongAnds6[1] & col) == wrongAnds6[1])
|| ((wrongOrs6[0] | col) == wrongOrs6[0])
|| ((wrongOrs6[1] | col) == wrongOrs6[1])
) {
toBottomOk = false;
}
}
if(toRightOk){
if(!toRight[grid1]){
toRight[grid1] = [];
}
toRight[grid1].push(grid2);
}
if(toBottomOk){
if(!toBottom[grid1]){
toBottom[grid1] = [];
}
toBottom[grid1].push(grid2);
}
}
}
function intersect(arr1, arr2){
var results = [];
for (var i = 0; i < arr1.length; i++) {
if (arr2.indexOf(arr1[i]) !== -1) {
results.push(arr1[i]);
}
}
return results;
}
var found = false;
while(!found){
var grid1 = threeGrid[0];
var grid1 = threeGrid[Math.floor(Math.random()*threeGrid.length)];
var grid2 = toRight[grid1][Math.floor(Math.random()*toRight[grid1].length)];
var grid3 = toBottom[grid1][Math.floor(Math.random()*toBottom[grid1].length)];
var arr4 = intersect(toBottom[grid2], toRight[grid3]);
if(arr4.length > 0){
var grid4 = arr4[Math.floor(Math.random()*arr4.length)];
found = true;
}
}
function gridToStrings(grid){
var rowS = [];
for(var i=0; i<3; i++){
rowS.push(toBin(((grid & wrongAnds[i]) >> (i*3))));
}
return rowS;
}
var grid1S = gridToStrings(grid1);
var grid2S = gridToStrings(grid2);
var grid3S = gridToStrings(grid3);
var grid4S = gridToStrings(grid4);
print(grid1S[0] + grid2S[0]);
print(grid1S[1] + grid2S[1]);
print(grid1S[2] + grid2S[2]);
print(grid3S[0] + grid4S[0]);
print(grid3S[1] + grid4S[1]);
print(grid3S[2] + grid4S[2]);
});
Theory:
Find all possible 3x3 grids
Find all possible left-to-right and top-to-bottom pairings
get 4 random grids to form the 6x6 grid
Implementation:
Represent 3x3 grids as 9bit integers. A 3x3 grid is wrong if there are 3 1s or 3 0s in it. This can be easily filtered with a couple bitwise operations.
Test the Cartesian product of these 3x3 grids (Compare every grid with every grid). Check if there are exactly 3 0s and 3 1s in all rows and columns (put the second grid right to the first grid to check 3 rows, and put it below the first grid to check 3 columns), and that there are no consecutive 3 0s or 1s.
get the top-left, top-right and bottom-right grids. Check if there is an available 4th grid that can go below the top-right grid and right to the bottom-left grid. If there is none, restart step 4, otherwise pick one.
A couple outputs:
011010
100101
001011
110100
101100
010011
110010
101100
010011
001101
100110
011001
001101
110010
010011
101100
110100
001011
Edit:
there is only 1120 solutions to this problem (jsFiddle). There are 2^36 ways to fill a 6x6 grid with 0s and 1s. If you used brute force (get a random 6x6 grid, then check if its right), that would mean an average ~61356676 (6.1*10^7) executions to find a correct solution. Even thought your method is somewhat faster (it can fail sooner if its not the last digit thats wrong), it might still be slow.
I think there are two problems with your code:
If oCount or zCount have become 3 there are no more assignments grid[i][j]=current if the random value is not acceptable. You get zeroes at these positions (to which the grid was initialized).
Near the right bottom there might not be any more valid solutions. You would have to undo previous assignments, i.e. you would need to do some kind of backtracking.
I would recommend starting with a valid solution and transforming this solution step by step according to random values for grid positions - but only if this is possible without breaking validity. If have prepared an example implementation:
public static void main(String[] args) {
int l = 6, w = 6;
Grid g = new Grid(l, w);
Random rd = new Random();
// initialize with checkerboard pattern (which is a valid solution)
for (int y = 0; y < l; y++) for (int x = 0; x < w; x++) g.arr[y][x] = (x ^ y) & 1;
// construct a valid grid by transformation of grids while preserving validity
for (int y = 0; y < l; y++) for (int x = 0; x < w; x++) {
int v = rd.nextInt(2), v2 = v ^ 1;
if (g.arr[y][x] == v) continue;
// try to modify current grid by exchanging values: 01/10=>10/01 or 10/01=>01/10
// (keep parts of the grid which have already been adapted to random values)
rotating: for (int y2 = y + 1; y2 < l; y2++) for (int x2 = x; x2 < w; x2++) {
if (g.arr[y2][x] == v && g.arr[y][x2] == v && g.arr[y2][x2] == v2) {
g.rotate(x, y, x2, y2);
// keep result if grid is still valid, undo otherwise
if (g.rotatedOk(x, y, x2, y2)) break rotating;
g.rotate(x, y, x2, y2);
}
}
}
g.printOn(System.out);
}
public static class Grid {
int l, w;
int[][] arr;
Grid(int l, int w) {
this.arr = new int[this.l = l][this.w = w];
}
void rotate(int x, int y, int x2, int y2) {
int v;
v = arr[y][x]; arr[y][x] = arr[y2][x]; arr[y2][x] = v;
v = arr[y][x2]; arr[y][x2] = arr[y2][x2]; arr[y2][x2] = v;
}
boolean rotatedOk(int x, int y, int x2, int y2) { // check after rotation
return okAt(x, y) && okAt(x2, y) && okAt(x, y2) && okAt(x2, y2);
}
private boolean okAt(int x, int y) { // check single position in grid
int v = arr[y][x];
if (count(x, y, -1, 0, v) + count(x, y, 1, 0, v) > 1) return false;
if (count(x, y, 0, -1, v) + count(x, y, 0, 1, v) > 1) return false;
return true;
}
private int count(int x, int y, int dx, int dy, int v) {
for (int n = 0; ; n++) {
x += dx; y += dy;
if (x < 0 || x >= w || y < 0 || y >= l || arr[y][x] != v) return n;
}
}
void printOn(PrintStream s) {
for (int y = 0; y < l; y++) { for (int x = 0; x < w; x++) s.print(arr[y][x]); s.println(); }
}
}
The problem with your approach is that you need a mechanism that handles when a new value can't be used because it follows two similar values, but the other value can't be used because it is under two other values. For example, say your grid has got this far:
101010
011010
00?
You would then need to slowly roll back positions and try different values.
The following code solves that problem using recursion:
import java.util.Random;
public class Main {
final int height = 6;
final int width = 6;
int[][] grid;
Random rd = new Random();
public static void main(final String[] args) {
Main main = new Main();
main.process();
}
private void process() {
// Create a grid that is 6 x 6
grid = new int[height][width];
for(int x = 0; x < width; x++) {
for(int y = 0; y < height; y++) {
grid[x][y] = -1;
}
}
recurseFillMatrix(0, 0);
}
private boolean recurseFillMatrix(final int x, final int y) {
// first, try putting a random number in the cell
int attempt = 1;
grid[x][y] = Math.abs(rd.nextInt()%2);
do {
if(isGridValid()) {
if(x == (width - 1) && y == (height - 1)) {
printGrid();
return true;
}
boolean problemSolved;
if(x == (width - 1)) {
problemSolved = recurseFillMatrix(0, y + 1);
} else {
problemSolved = recurseFillMatrix(x + 1, y);
}
if(problemSolved) {
return true;
}
}
attempt++;
grid[x][y] = 1 - grid[x][y];
} while(attempt <= 2);
grid[x][y] = -1;
return false;
}
private boolean isGridValid() {
for(int y = 0; y < height; y++) {
for(int x = 0; x < width; x++) {
// if the current item is -1, then we are finished
if(grid[x][y] == -1) {
return true;
}
// if we are after the second column
if(x > 1) {
if(grid[x-2][y] == grid[x-1][y] && grid[x-1][y] == grid[x][y]) {
return false;
}
}
// if we are after the second row
if(y > 1) {
if(grid[x][y-2] == grid[x][y-1] && grid[x][y-1] == grid[x][y]) {
return false;
}
}
// total the values in this column
int total = 0;
for(int i = 0; i <= y; i++) {
total += grid[x][i];
}
if(y == (height - 1)) {
if(total != 3) {
return false;
}
} else {
if(total > 3) {
return false;
}
}
// total the values in this row
total = 0;
for(int i = 0; i <= x; i++) {
total += grid[i][y];
}
if(x == (width - 1)) {
if(total != 3) {
return false;
}
} else {
if(total > 3) {
return false;
}
}
}
}
return true;
}
private void printGrid() {
for(int y = 0; y < height; y++) {
for(int x = 0; x < width; x++) {
System.out.print(grid[x][y]);
}
System.out.println("");
}
}
}
The isGridValid() method uses your defined rules to check if the grid (as it is filled so far) complies with the rules. At the first sign that it does not, it returns false.
If I have to change your solution to achieve the result, here is what it should look like..
Take the incrementors for oCount and zCount in a separate if-else
Take the assignment to grid(i,j) outside the loop
Your if-else block is not taking into account every condition possible, like
What about when last 2 items are same
What about when the zCount or oCount has reached 3
Taking into account these consideration, this code works fine.
import java.util.Random;
public class Main {
public static void main(String[] args) {
int l = 6;
int w = 6;
Random rd = new Random();
// Create a grid that is 6 x 6
int[][] grid = new int[l][w];
// for each row
for (int i = 0; i < l; i++) {
int zCount = 0;
int oCount = 0;
int current;
int lastA = 2;
int lastB = 2;
// for each item in the row
for (int j = 0; j < w; j++) {
// set the current item to either 0 or 1
current = rd.nextInt(2);
// make sure there aren't already (e.g. 3 items out of 6)
// items in the row
if (current == 1) {
if (oCount != 3) {
if (lastA == lastB) {
current = lastA == 1 ? 0 : 1;
}
} else {
current = current == 1 ? 0 : 1;
}
} else if (current == 0) {
if (zCount != 3) {
if (lastA == lastB) {
current = lastA == 1 ? 0 : 1;
}
} else {
current = current == 1 ? 0 : 1;
}
}
grid[i][j] = current;
if (current == 1) {
oCount++;
} else {
zCount++;
}
if (j % 2 == 1) {
// hold every second element
lastA = current;
} else {
// hold every first element
lastB = current;
}
System.out.print(grid[i][j]);
}
System.out.println(" ");
}
}
}
Again, This solution takes care of row conditions only. You would need to do similar checks for columns as well, to achieve the full result
HTH
here I tested you problem and seems that it is what you need.
I used a functional approach using Guava, it is quite simple, readable and has a short code.
#Test
public void test_permutations()
{
List<Integer> binary = Lists.newArrayList(1,0,1,0,1,0); // Domain list
Set<String> flattenSet = Sets.newHashSet(); // Store non-repetitive values
// Create list of possible values
Collection<List<Integer>> permutations = Collections2.permutations(binary);
for (List<Integer> permutation : permutations)
{
String joinString = StringUtils.join(permutation, "");
flattenSet.add(joinString);
}
// Create predicate to filter positive values
Predicate<String> predicate = new Predicate<String>() {
public boolean apply(String input) {
// Discard wrong values
if (input.contains("000") || input.contains("111")) {
return false;
} else {
return true;
}
}
};
// Use predicate to filter values
Collection<String> filteredList = Collections2.filter(flattenSet, predicate);
// Display result
for (String result : filteredList) {
System.out.println(result);
}
}
It is simple, I've commented the code to be clear but you can debug it to understand step by step.
The generated output is:
010011
110010
010101
010110
100110
101001
011010
110100
001011
001101
011001
101010
101100
100101
Hope to help
I think that its a mistake to think of generating it one element at at time. Instead imagine that I generate the entire set of permissible rows {001100,101010,....etc} There are only 6!/(3!3!)=20 ways to arrange three ones and three and some of them will be excluded. Now I am going to generate a game tree by saying that a move is selecting a valid row for the next row. If I discover at some point that there are no more valid moves then i will back track and try a different move.
To generate a move I randomly select a row, if its a valid move, I try to select another move, if that is impossible I backtrack, effectively doing a (random) depth first search of the game tree.
public class gametree {
public static ImmutableList<Row> allValidRows = // create a list of all valid rows.
public static List<Rows> getValidMoves(Move parent){ //Backtracks up the
//tree to the root to find the current state of the board, and returns
//which ever of allValidRows are valid given the game board.
}
public class Move {
public final Move parent;
public List<Rows> validMoves;
public final Row thisMove;
public int depth=0;
Move(Move parent, Row thisMove){
this.thisMove = thisMove;
this.parent = parent;
this.validMoves = getValidMoves(parent);
Move hold=parent;
while(hold!=null){
depth++; hold = parent.parent;
}
}
}
void run {
//pick first move
Move Root = new Move(null, Collections.Shuffle(allValidRows).get(0));
Move FinalMove = search(Root);
//Something to print out the answer here
}
public Move search(Move move){
if(depth==5){ return Move} //If I get to row six I win.
else if(move.validMoves.isEmpty()) { //If there are no valid moves,
//then this move wasnt valid, to strip it from the parent's
//possible moves and try again
move.parent.validMoves.remove(move.thisMove);
search(move.parent);
} else { //pick a random valid move and create a nextMove
Move nextMove = new Move(move, Collection.Shuffle(move.getValidMoves).get(0))
search(nextMove);
}
}
The worst case for this algorithm is that there is only one victory state and it has to try every possible state, but in practice this game does not seem very restrictive so it will probably not take long at all.
This code is strictly illustrative.